|f 430 Moment of Resistance. t; 1 p We may also represent the moment in terms of the limiting !! stress/(sometimes/c, and sometimes/t). Then: if Jj Bending moment = moment of resistance j| Bm=/Z ........................ (4) II I/ and Z is known as the modulus of section,* ^ || Let y = distance of furthest fibre from axis: I By (i) /= by (3) Bm = — and by (4) Bm=/Z Z-— .-. Z = *...... (5) p P ° y and Bm=/- . _____I (Seej). 1079.) The value of Z can now be found; thus, p 1 % f r i • rr, r " '" ^ f b frfi rectangular Sections /Z =/------r - =/-— and for circular sections /Z = /—— -r- - ----------------------------£---- J 64 2, Graphic Solution for Moment of Resistance.— Taking, first, a rectangular section (Fig. 384), draw the neutral axis A B. Then c D will be the line of limiting (or greatest) stress, and the value of any horizontal fibre E F to resist stress will be found by projecting to c D and joining c D to N, thus obtaining the intercept G M. Every fibre being thus treated, the sum of the -virtual stress areas will be the areas c{# N and H j N, which each make one force of the couple when multiplied by the limiting stress f. K and L are the centres of gravity of the areas. Moment of resistance (generally) = one force x total arm Moment of rectangular section = f\area C D N [• x arm K L Unsymmetrical sections are treated at Figs. 389, 390 by this method, which can be applied to any section. (SeeApp.IL>p. 847.) •* Z = virtual area x arm. (See Figs. 384, 385.)